Accelerometer placement – where and why

A question that should be asked during physical design of any product incorporating sensors is: “Where should be sensors be located within the product?” Like almost anything else in life, “It depends … ” is the answer. This post discusses accelerometer placement. We’ll touch on magnetic sensors next time around.

If your application is such that the sensor is statically fixed in space, or at least not rotating, it doesn’t really matter where you mount the accelerometer on your board. But if the device can rotate, things get a lot more interesting.

The rate of change transport theorem, which can be found in any rigid body dynamics textbook, relates the rate of change of a vector r as observed in two different reference frames. For our purposes, the first is the fixed (earth) frame and the second (body frame) rotates and translates relative to the first. In equation form:

(dr/dt)f =(dr/dt)r + ω X rr (Eqn. 1)

where the subscripts indicate the frame of reference.

The transport theorem is a fundamental relationship, based upon basic principles of geometry. It applies to ANY vector observable from two independent reference frames. You should note that both left and right sides of the equation are vectors.

We are going to repeatedly apply the transport theorem to derive a relationship defining the acceleration of an object in the fixed earth frame based upon sensor measurements made in the body frame.

In graphical form, consider movable point “P”, observable from two different frames of reference. “Of” is the origin of the fixed earth reference frame. “Or” is the origin of the rotating reference frame.

Frames of Reference

By basic geometry

r’ = R + r (Eqn. 2)

Note that

V = (dr/dt)f

V is the rate of change of vector r as viewed in the fixed reference frame. It can also be viewed as the velocity of P as viewed in the fixed earth frame

Vr = (dr/dt)r

Vr is the he rate of change of vector r as viewed in the rotating reference frame. It can also be viewed as the velocity of P as viewed within the rotating reference frame.

ω = the angular velocity of the rotating frame relative to the fixed earth frame

by differentiating Eqn. 2, we get

(dr’/dt)f = (dR/dt)f + (dr/dt)f (Eqn. 3)

which we can also write as

Vr’ = V + (dr/dt)f (Eqn. 4)

The quantity V can be interpreted as the translational velocity of the rotating frame relative to the fixed frame of reference. Vr’ is the velocity of point P as observed in the fixed earth frame. Typically, this is the quantity we are trying to measure.

We can apply the transport theorem to the right most term in Eqn. 4 to get

where A is the acceleration of the second frame of reference relative to the fixed earth frame.

At the same time, apply the transport theorem again to substitute

(dVr/dt)f =(dVr/dt)r+ ω X Vr (Eqn. 9)

to get

Ar’ = A+ (dVr/dt)r+ ω X Vr+ (dω/dt)f X r + ω X (dr/dt)f (Eqn. 10)

Let

α = (dω/dt)f

where α is the rate of angular acceleration, and

Ar = (dVr/dt)r

where Ar is the rate of acceleration of P relative to Or. Then

Ar’ = A+ Ar+ ω X Vr+ α X r + ω X (dr/dt)f (Eqn. 11)

and one last substitution using the transport theorem for the right most term in Eqn. 11 to get

Ar’ = A+ Ar+ ω X Vr+ α X r + ω X ((dr/dt)r + ω X r) (Eqn. 12)

Ar’ = A+ Ar+ ω X Vr+ α X r + ω X (Vr + ω X r) (Eqn. 13)

Ar’ = A+ Ar+ ω X Vr+ α X r + ω X Vr + ω X (ω X r)) (Eqn. 14)

Ar’ =

A + Ar +

<– Inertial Acceleration

2ω X Vr +

<– Coriolis Acceleration

α X r +

<– Euler Acceleration

ω X (ω X r))

<– Centripetal Acceleration

(Eqn. 15)

Eqn. 15 is a classic result, and can give you a lot of insight into tradeoffs associated with accelerometer placement. To see how, let’s modify our previous figure by locating Or at the center of mass of a portable consumer device. Point P represents the location of our accelerometer within the same device. Point Of is essentially the center of the earth.

Frames of Reference related to Portable Consumer Device

Since points P and Or are fixed relative to one another (they are part of the same rigid body), Ar and Vr are zero by definition, and drop out of the equation. If the rotational rate of the device is zero, then ω=0, and both Coriolis and Centripetal acceleration terms drop out of the equation. Finally, if α (the rotational acceleration) is zero, there will be no Euler acceleration. If all these terms hold true, then the value measured by our accelerometer is

Ar’ = A = the translational rate between the two frames of reference

which is exactly what we would like. But what if ω≠0? If we collapse point P back into the center of mass (Or), then both Euler and Centripetal accelerations still drop out.

What happens if you ignore these effects? Let’s work an example assuming that angular acceleration is zero, but angular rotation is not:

Ar’ =

A + Ar +

<– Inertial Acceleration

ω X (ω X r))

<– Centripetal Acceleration

(Eqn. 16)

To bound the problem, I’m going to take advantage of the identity:

A X B = |A| |B| sin Θ n

where Θ is the angle between vectors A and B, and n is a unit vector perpendicular to the plane containing both A and B.

Looking at just the extremes, we can see that the Centripetal acceleration ranges between +/- |ω|2r in magnitude, depending upon the axis of rotation and accelerometer offset from center of mass.

Assume

ω = 2π radians/sec (1 RPM)

r = 5 cm

Putting numbers to Eqn. 16, we get

Centripetal Acceleration = +/- 0.05 (2π)2 =+/- 1.97 m/s2.

That’s better than 1/5 gravities! At a very modest rate of rotation!

Basically, sensor fusion mathematics get much easier when the accelerometer is at the center of mass (which is also the center of the rotating frame of reference). If you can make the assumption that, in your use case, the device is in a quasi-static state with no rotation, it doesn’t matter. If not, you need to place the sensor at the center of mass OR account for the effects mathematically. That involves keeping continuous track of the axis and rate of rotation and then making the adjustments required to separate out the Centripetal acceleration.